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\title{Operating Systems - assignment 4}
\author{Camil Staps\\\small{s4498062}}

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\section*{6.11}
\begin{enumerate}
    \item No process is marked, because no process has all zeros in the allocation matrix.
    \item $W \coloneqq (2,1,1,2)$.
    \item Take $i=1$.
    \item Mark $P_1$, $W \coloneqq (2,1,2,2)$.
    \setcounter{enumi}{2}
    \item Take $i=3$.
    \item Mark $P_3$, $W \coloneqq (2,2,3,2)$.
    \setcounter{enumi}{2}
    \item Both $Q_2 > W$ and $Q_4 > W$, so we terminate.
\end{enumerate}

There are unmarked processes ($P_2,P_4$), therefore, deadlock exists, for exactly those processes.

\section*{6.18}
\begin{enumerate}[label=\alph*.]
    \item If there are two types of philosophers, the fourth condition of deadlock (circular wait) is not fulfilled. Lefties and righties can never be in a chain waiting for resources. Therefore, deadlock cannot occur.
    \item On a table with two philosophers, of which one is a lefty and one a righty, it is obvious to see that one philosopher will start eating and when he is done, the other one can grab the forks. No starvation occurs.

        If on a table with $k$ philosophers no starvation occurs, and we add another philosopher $p$, this new philosopher is placed between two philosophers who, at some point, will eat. Then at some point the first fork $p$ wants to take will be free, and also at some point the other fork $p$ needs will be free. Hence, $p$ will also eat.

        From the principle of induction it now follows that starvation cannot occur on a table with $n\ge2$ philosophers, as long as there is at least one lefty and one righty.
\end{enumerate}

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